A herd of bison was placed in a wildlife preserve that can...

A herd of bison was placed in a wildlife preserve that can support a maximum of 1000 bison. A population model for the bison is given by $$B=\frac{1000}{1+30 e^{-0.127}}$$ where $B$ is the number of bison in the preserve and $t$ is time in years, with the year 1999 represented by $t=0$ a. Use a graphing utility to graph the equation for $t \geq 0$ b. Use the graph to estimate (to the nearest year) the number of years before the bison population reaches $500 .$ c. Determine the horizontal asymptote of the graph. d. Write a sentence that explains the meaning of the horizontal asymptote.

A herd of bison was placed in a wildlife preserve that can support a maximum of 1000 bison.
A population model for the bison is given by
$$B=\frac{1000}{1+30 e^{-0.127}}$$
where $B$ is the number of bison in the preserve and $t$ is time in years, with the year 1999 represented by $t=0$
a. Use a graphing utility to graph the equation for $t \geq 0$
b. Use the graph to estimate (to the nearest year) the number of years before the bison population reaches $500 .$
c. Determine the horizontal asymptote of the graph.
d. Write a sentence that explains the meaning of the horizontal asymptote.

Key Concepts

Logistic Growth Model

A logistic growth model is used to describe populations that grow rapidly at first and then slow as they approach a maximum sustainable size, known as the carrying capacity. This model produces an S-shaped, or sigmoidal, curve that takes into account environmental limitations and resource constraints.

Carrying Capacity

The carrying capacity is the maximum number of individuals that an environment can sustainably support. In logistic models, the carrying capacity is reflected as the horizontal asymptote that the population approaches as time increases.

Horizontal Asymptote

In the context of logistic functions, the horizontal asymptote represents the limiting value that the function approaches as the independent variable (typically time) goes to infinity. This asymptote indicates the long?term behavior of the system and, in population models, corresponds to the carrying capacity.

Exponential Decay Factor in Logistic Models

The exponential decay term in a logistic growth equation, such as e^(–0.127t), controls the rate at which the population transitions from its initial size toward the carrying capacity. It helps model the initial phase of rapid growth and the subsequent slowdown as the population nears its environmental limit.

Graphical Analysis for Estimation

Graphical analysis involves plotting the logistic function and visually identifying features such as the intersection with a horizontal line representing a target population size. This method is useful for estimating key points, such as the time needed for the population to reach a specific level, and for understanding the overall dynamics of the growth model.

Transcript

00:01 Talk about question number 63.

00:03 So we are given that a herd of bison was placed in a wildlife preserve that can support a maximum 1 ,000 bison.

00:12 Population model of the bison is given by this equation where b is the number of bison is the in the preserve b is the number of bison in the preserve and t is the time in years with the year 1999 represented as t equal to zero it means that the t equal to one represents 2000 t equal to two represents 2001 and so on we need to we need to use a graphical utility to first graph this particular equation so let's do that we have 1 ,000 over 1000 is called it voy as equal to 1000 over 1 plus 30 times e uh raised to minus 0 .1 to 70 so that's minus 0 .127.

01:00 And this we have to graph for the positive values of t.

01:05 So this is how the graph looks like.

01:11 It starts from 32 .258 and it goes till 1000.

01:19 Then we have to use the graph to determine the nearest year.

01:24 When the number the number of years to the nearest year when the bison population reaches 500 so it means that we need to solve for so part a is already done part b is we need to solve b is replaced by 500 so we need to solve for 500 is equal to 1000 over 1 plus 3 times e raised to negative 1 .27 negative 0 .1 270 so another equation is y is equal to 500 and we need to look for its point of intersection.

01:59 So clearly this is the point of intersection.

02:02 The value of t is 26 .78.

02:05 So to the nearest year, that's 27...

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